[英]await async lambda in ActionBlock
我有一个带有ActionBlock的类Receiver:
public class Receiver<T> : IReceiver<T>
{
private ActionBlock<T> _receiver;
public Task<bool> Send(T item)
{
if(_receiver!=null)
return _receiver.SendAsync(item);
//Do some other stuff her
}
public void Register (Func<T, Task> receiver)
{
_receiver = new ActionBlock<T> (receiver);
}
//...
}
ActionBlock的Register-Action是一个带有await-Statement的async-Method:
private static async Task Writer(int num)
{
Console.WriteLine("start " + num);
await Task.Delay(500);
Console.WriteLine("end " + num);
}
现在我想做的是同步等待(如果设置了条件),直到action方法完成以获得独占行为:
var receiver = new Receiver<int>();
receiver.Register((Func<int, Task) Writer);
receiver.Send(5).Wait(); //does not wait the action-await here!
问题是“await Task.Delay(500);” 语句被执行,“receiver.Post(5).Wait();” 不再等了。
我尝试了几种变体(TaskCompletionSource,ContinueWith,...),但它不起作用。
有谁知道如何解决这个问题?
默认情况下, ActionBlock
将强制执行独占行为(一次只处理一个项目)。 如果您通过“独占行为”表示其他内容,则可以使用TaskCompletionSource
在操作完成时通知您的发件人:
... use ActionBlock<Tuple<int, TaskCompletionSource<object>>> and Receiver<Tuple<int, TaskCompletionSource<object>>>
var receiver = new Receiver<Tuple<int, TaskCompletionSource<object>>>();
receiver.Register((Func<Tuple<int, TaskCompletionSource<object>>, Task) Writer);
var tcs = new TaskCompletionSource<object>();
receiver.Send(Tuple.Create(5, tcs));
tcs.Task.Wait(); // if you must
private static async Task Writer(int num, TaskCompletionSource<object> tcs)
{
Console.WriteLine("start " + num);
await Task.Delay(500);
Console.WriteLine("end " + num);
tcs.SetResult(null);
}
或者,您可以使用AsyncLock
( 包含在我的AsyncEx库中 ):
private static AsyncLock mutex = new AsyncLock();
private static async Task Writer(int num)
{
using (await mutex.LockAsync())
{
Console.WriteLine("start " + num);
await Task.Delay(500);
Console.WriteLine("end " + num);
}
}
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